## PSN-L Email List Message

From: "Stephen Hammond" shammon1@.............
Date: Sun, 6 Dec 2009 22:58:13 -0800

```Hi Chris, you got my attention when you talked about the length of the =
arm
and mass.  Let=92s say A=92 is the pipe end of the arm, B=92 is the =
location of
the mass (say 5 lbs) and =93C=94 is the location of the brass damping =
plate ( =BC
lbs) on the original Lehman style design. Using your calculation below,
would you measure the 24=94 arm from A=92 to B=92 or A=92 to C=92 ?  =
Take a look at
http://pw2.netcom.com/~shammon1/compressed/e-w_Lehman.jpg and you will =
see
why I am asking. When I was in San Jose I had two Lehman=92s with 36=92 =
arms and
the worked great and they had 20-25 second periods without any effort. =
When
I moved to Aptos I had to reduce the length to 20=94 due to space issues =
and
they have never performed as well as I predicted. Now I=92m wondering if =
I am
measuring their arm length correctly.   Comments?

Steve Hammond PSN Aptos, CA

=20

From: psn-l-request@.............. [mailto:psn-l-request@............... =
On
Behalf Of ChrisAtUpw@.......
Sent: Sunday, December 06, 2009 9:28 AM
To: psn-l@..............

=20

In a message dated 06/12/2009, tchannel@............ writes:

Hi Folks,   This is the basic process I use to adj. a Lehman.

1.  I assembly everything and level everything.    Once this is =
completed
the vertical part of the frame is plumb.   The boom, and frame are =
parallel
with the earth and the boom is resting on the centreline.

2.  I time the period and start raising the front adjusting leg, =
tweaking
the side adj. legs as necessary to maintain the boom on the centre line.
When I get a stable 20 second period I am finished.

=20

These are my questions:    When I perform this set up, the vertical =
frame
member is now tilting backward, and the mass end of the boom has be =
raised.
Now the boom is no longer parallel to the earth.

1.  "What would happen" if I extended the wire support, lowering the =
boom
back to parallel to the earth, but not changing the vertical member.   =
To my
knowledge I have never done this, I just leave the boom with mass end =
raised

2.  Should I make this adjustment, returning the boom to parallel, by
lengthen the support wire?

I have tried it, but the adjustments are so small I can't measure the
results.

Hi Ted,

=20

The period is determined by 1) the angle 'A' between the vertical =
and
the true swing axis and by 2) the radius of gyration 'k' of the =
pendulum. k
depends on the weight and shape of the mass and of the arm. It is usual =
to
use a rigid light tubular arm. Lifting or lowering the arm a bit will
provide a Cosine Theta correction to k, so you could see a few parts per
thousand change.=20

=20

Usual values of the swing angle are about 1/3 degree for a 20 second
period and a 24" arm. T =3D 2xPixSqrt(k / gxsin(A)) Avoid very short =
arms, say
12". The value of A becomes very small and the pendulum will experience =
tilt
drift due to very small ground movements as the seasons, rainfall and
temperatures change. You need to design your sensor system so that it's
output is independent of any tilt drift.

The layout of the sensor and the magnetic damping with the frame =
should

=20

Regards,

=20

Chris Chapman

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Hi Chris, you got my attention when you talked about the =
length
of the arm and mass.=A0 Let’s say A’ is the pipe end of the =
arm, B’
is the location of the mass (say 5 lbs) and “C” is the =
location of
the brass damping plate ( =BC lbs) on the original Lehman style design. =
Using
your calculation below, would you measure the 24” arm from =
A’ to B’
or A’ to C’ ?=A0 Take a look at http:/=
/pw2.netcom.com/~shammon1/compressed/e-w_Lehman.jpg
and you will see why I am asking. When I was in San Jose I had two =
Lehman’s
with 36’ arms and the worked great and they had 20-25 second =
periods
without any effort. When I moved to Aptos I had to reduce the length to =
20”
due to space issues and they have never performed as well as I =
predicted. Now I’m
wondering if I am measuring their arm length correctly.=A0=A0 =

Steve Hammond PSN Aptos, CA

From:=

psn-l-request@.............. [mailto:psn-l-request@............... On =
Behalf
Of ChrisAtUpw@.......
Sent: Sunday, December 06, 2009 9:28 AM
To: psn-l@..............

In a message dated 06/12/2009, tchannel@............ =
writes:

Hi Folks,   This is the basic process I use to =
Lehman.

1.  I assembly everything and level
everything.    Once this is completed the vertical part =
of the
frame is plumb.   The boom, and frame are parallel =
with the
earth and the boom is resting on the centreline.

2.  I time the period and start raising the front =
leg, tweaking the side adj. legs as necessary to maintain the boom on =
the
centre line.  When I get a stable 20 second period I am =
finished.

These are my questions:    When I =
perform this
set up, the vertical frame member is now tilting backward, and =
the mass
end of the boom has be raised.  Now the boom is no longer parallel =
to the
earth.

1.  "What would happen" if I extended the =
wire
support, lowering the boom back to parallel to the earth, but not =
changing the
vertical member.   To my knowledge I have never done this, I =
just
leave the boom with mass end raised via the front adj. =
leg.

2.  Should I make this adjustment, returning the boom =
to
parallel, by lengthen the support wire?

I have tried it, but the adjustments are so small I can't =
measure
the results.

Hi Ted,

The period is determined by 1) the =
angle
'A' between the vertical and the true swing axis and by 2) the =
gyration 'k' of the pendulum. k depends on the weight and shape of the =
mass and
of the arm. It is usual to use a rigid light tubular arm. Lifting =
or
lowering the arm a bit will provide a Cosine Theta correction to k, so
you could see a few parts per thousand =
change.

Usual values of the swing angle are =
1/3 degree for a 20 second period and a 24" arm. T =3D 2xPixSqrt(k =
/
gxsin(A)) Avoid very short arms, say 12". The value of A becomes =
very
small and the pendulum will experience tilt drift due to very small =
ground
movements as the seasons, rainfall and temperatures change. You =
need to
design your sensor system so that it's output is independent of any tilt =
drift.

The layout of the sensor and the =
magnetic
damping with the frame should be your primary =
concern.

Regards,

Chris Chapman

```